JAMB - Physics (2025 - No. 14)
A Force 18 N pulls a 40 kg mass on a horizontal floor at 0.3 ms\(^{-2}\). Find the coefficient of friction.
0.003
0.015
0.300
0.600
स्पष्टीकरण
Answer: B. 0.015
(Net F = ma = 40×0.3 = 12 N;
friction = 18 - 12 = 6 N;
μ = \(\frac{\text{friction}}{\text{mg}}\) = \(\frac{6}{(40×10)}\) = 0.015.)